# Real Steel Greenhouse Effect

13 years ago, amateur scientist Willis Eschenbach developed a thought experiment that he hoped would very simply illustrate how the Greenhouse Effect works.

The main claim is that the addition of a steel shell surrounding a planetary surface will cause the inner surface to emit TWICE (235 -> 470 W/m2) the radiation as compared to not having steel shell. This should significantly raise the surface temperature from ~254K to ~302K.

Is this true? No.

Willis gives us the freedom to construct any power source with any chosen radius. I will choose a mini nuclear reactor wrapped in a steel housing, with the total surface area being 1 m2. The inner radius of steel housing is 75% of total radius.

Let us go through the equations to make set up Willis’ initial scenario (A):

The nuclear power reactor is ONLY capable of making its wall 254.041K – to meet Willis’ initial criteria. It is not capable of anything greater, because nuclear reactions are fixed. No varying levels of downstream radiation will enable nuclear fission reactions to generate more joules.

Now let us see what happens when we add a steel shell (B):

I will give Willis credit for doing a good job of demonstrating the real greenhouse effect:

Outgoing radiation is halfed and T2 (our “surface”) has increased from 253.726K to 253.884K, a very feeble gain.

The problem with Willis’ approach is that he doesn’t reduce outgoing radiation and relies on his heat source to crank up … when there is no physical way it can do so.

Summary:

So there you have it. The real steel greenhouse effect managed to raise the surface temperature by 0.062%.

Subsequent additions of steel shells will keep raising the surface temperature (T2), in an inverse asymptotic fashion approaching nuclear reactor wall temperature (T1).

Enjoy 🙂 -Zoe

https://phzoe.com

## 132 thoughts on “Real Steel Greenhouse Effect”

1. Zoe, I can’t check your maths. Let’s hope Willis shows up to defend his argument. Intuitively, steel is a good conductor. But a steel shell would stop the atmosphere convecting. Convection elevates a molecule to a height where there is less material that is capable of absorbing radiation directly to space. Ozone is a large molecule that efficiently absorbs wave lengths that are most commonly emitted by the Earth system and its presence gives rise to an increase in gas temperature above an altitude of 10,000 metres in the mid latitudes. There is no evidence that the enhanced emission by the molecules warmed by ozone (nitrogen and oxygen) actually warms the layers beneath the point where that temperature increase occurs.

The supposed warming effect of CO2 fails to take into account the efficiency of convective processes.

Willis should spend a little time living in a sea container in an environment where air temperature falls to near zero overnight.

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1. Well, um, there is no atmosphere in this example. No convection at all.

Merely adding conductive matter on top of our “surface” would also raise temperature, but not as much as a detached shell via radiative means.

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2. Max Polo says:

Thanks, Zoe, so interesting as always. I plan to check your math soon 😉

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1. Ha ha. Please do. Computer formatting formulas is a pain in the neck.

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3. thecliffclavenoffinance says:

I don’t think it’s fair to call Willis an amateur scientist,
nor would it be fair to use that title for you.
You both present scientific articles, so are scientists
by my definition of scientist, even if not paid for your work.
There are plenty of people with science degrees
paid to do reckless climate scaremongering. They
are called “scientists”, but always wrong wild guesses
of the future climate are not science.

No need to print the following
portion of my comment:

Your three posts in three days is like a miracle.
Every reader wishes you would post more often.
Once in a week, or two, would be fine
if you had the time. It’s nice to be “in demand”.

I publish a climate science and energy blog
where I present the best articles I’ve read that
day. So far over 319,000 page views.
I have only recommended two other blogs
in the past few years, that have consistently
good articles. Your blog is also consistently
good, and I’d recommend it … but …
you don’t have a regular publishing
schedule, so if people visit on my
recommendation, and see nothing new
for a month or two, they’ll get mad at ME.
This already happened with two friends
global sea ice area articles
— both of them thought you had
after those two articles.

If you had a regular schedule, such as a
a post every week, or every two weeks,
or even once a month, that would allow me
I get 300 to 400 page views a day.

PS: If you are anything like my wife,
you are now mad at me for giving
on Willis E. ha ha

Richard Greene
Bingham Farms, Michigan
http://www.elOnionBloggle.Blogspot.com

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1. I’m not mad. I’m a just very busy gal. I’d love to have more posts, but, as you can see, I do original research. Original research is very time consuming. When I don’t have the time, I make errors.

I don’t want to have a blog that just links to other articles. Neither do I want to have boiler plate posts about updates to climate data.

Right now I have just 2 ideas for next posts. After that, I will again have to scour for something original.

Ideally I’d like to share everything I know, but I think that will be boring for most people. My computer skills aren’t interesting to people. Those posts get like no views.

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1. thecliffclavenoffinance says:

So, let me annoy you even more with
an example. Let’s say you have the time
would be happy if they knew Zoe would
be publishing something in the first week
of every month. That would mean NOT
publishing three articles in three days,
as your just did. Those three articles could
have been published over three months.
When readers don’t see a post for a long
time, or there is no predictable schedule,
they stop looking. … I know genius can’t
be rushed, but if you publish three articles
in three days, and perhaps nothing else
for the next month, readers disappear.
That’s just my two cents, which is
probably worth one cent after inflation.

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1. This is not a job. I’m not on a schedule. When I have free time and desire, I write articles. Making this a scheduled task is making it like work, of which I have plenty.

Thanks for the suggestion tho. Let’s see what happens.

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2. thecliffclavenoffinance says:

Zoe:
Sorry, I jumped to the conclusion that
That’s apparently not a priority.
I also failed to explain my ideas well.
There is no need for you to do more work,
or to work at different times.
have to match your writing schedule.
Your work is just as valuable if presented
the day it is finished, or many months
later.

friends in late 2021 after reading two articles
on Global Sea Ice Area. With no new articles
in the next 2+ months, both friends deleted
your blog from their Bookmarks list, and never
visited again. Which I thought was a shame.

My own blogs serve a different purpose.
I merely share the best articles I’ve read
that day, and rarely write my own articles.
I used to write articles for my for-profit
1977 to 2020. That was a lot of work.

Now I use my blogs as an antidote
for social and mass media censorship,
and parroting government press releases.
So I want more readers, and assumed that
desire applied to everyone with a blog
or website.
I was just trying to be helpful.
Richard Greene
Bingham Farms, Michigan

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3. Nah. Publishing is the instant reward for writing. I have 7 posts I never published. I’m still writing them. lol. Actually they are crap, and I should delete them. And that happens every time I take that approach. If I start writing, it must be finished today or early tomorrow, or never.

Don’t apologize. Thank you for the advice.

I didn’t even think anyone would read my blog.

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4. Jarle says:

So the outgoing radiation is halved, while the source output stays the same. Please explain how T2 remains largely unchanged. I follow the math, but to me it is counter intuitive.

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1. What were you expecting? Willis result or no warming at all?

The net flow became ~118W, down from 235W. Less cooling … warmed up.

A net flow of 0W between surface and shell would mean the surface and shell is same temperature as nuclear reactor wall. This is impossible but presented for guidance.

Nothing in this system will exceed nuclear wall temperature.

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1. Jarle says:

I was expecting no warming. To me it is intuitive that no part of the system can achieve a higher temperature than the nuclear wall temp. It is also intuitive that the source output remains constant. So I guess I was expecting the cooling to remain the same, but after some condideration I agree that it must reduce. Radiation is inefficient

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1. Just to clear things up … I never doubted the REAL “greenhouse effect” (really: thermal resistence effect) of preventing cooling leading to a warming. I just never believed warming can exceed the SOURCE. I still don’t.

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2. The greenhouse effect depends upon stability in the medium. The atmosphere is anything but stable. Its highly mobile. Cooling ultimately depends on radiation to space but in the interim, it is very efficiently maintained via conduction and displacement, ie convection. The part of the globe that receives the most intense radiation, inside the tropics of Capricorn and Cancer, is temperature limited due to the efficiency of plant transpiration, evaporation from open bodies of water, the movement of ocean currents and most of all, convection. Accordingly, the tropics radiate much less energy to space than is received directly from the sun due to the temperature decline as gas expands as it is elevated. It is here that the tropopause, where the air is coldest, due to simple decompression and the impact of convection in displacing ozone upwards, that the air is as cold as over the poles in winter.

As the Hadley cell is energized by convection, the descent and compression of air in the mid latitudes is enhanced. Here, the air is dry, its greenhouse potential reduced and radiation to space proceeds with relative freedom. Dry air is not conducive to the production of cloud. The temperature of the air in the mid latitudes, is dependent, as it is everywhere else, on the partial pressure of ozone that is elevated by convection in extratropical (polar) cyclones to the upper limits of the atmosphere, from where it is transported to the mid latitudes, affecting cloud cover.

This is a complex system, incapable of being mathematically modelled. The most sophisticated minds think in terms of an ‘open system’ whereby the atmosphere is susceptible to change according to change in the wider solar and intergalactic influences. Sophistication is gained via a study of the geography and history of the atmosphere.

Greenhouse effect. Propaganda. A simple idea, easily assimilated, that is maintained in order to secure an economic, social and political objective. A scam.

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5. Jarle says:

The greenhouse effect is what it is. The claim that CO2 is a control button for the climate i something different. Most scientists do not dare to challenge this view, in fear of destroying their own reputation. Roy Spencer is known for having said that since the climate system is so complicated, we do not know whether humans have caused 10% or 90% of the observed warming. Of late though, he feels obliged for some reason, to state that: “I still provisionally side with the view that warming has been mostly human-caused”. He can’t back it up, but safe to preserve ones reputation since there is doubt. And this is how it goes, decade after decade. I hope this is the last decade. The truth must come, be it in favor of the alarmist or the sceptic side.

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1. CO2 causes between 0% and 100%
of the global warming, with both
0% and 100% unlikely, in my opinion.

I gave Roy Spencer a hard time on his blog
when the correct answer is no one knows.
I liked his 10% to 90% range.
But then I felt bad when later found out UAH
had been defunded and he and Christy
were providing UAH data on a voluntary basis
for the past few years. Hard to criticize a
scientist who does that.

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6. Steve Titcombe says:

Firstly, Willis Eschenbach’s thought experiment does NOT reflect how a real greenhouse works and neither does it reflect how the man-made (fictional) climate change greenhouse effect works – as these both require an atmosphere, whereas the thought experiment requires a vacuum within the infinitely small gap between the inner sphere and the outer shell. However, the thought experiment DOES correctly demonstrate how a colder object (the outer steel shell) can make a hotter object (the inner sphere) even hotter. I admit to once being a Joe Postma supporter in believing that a colder object could never make a hotter object even warmer – but I was wrong.

The thing that everyone who refutes Willis Eschenbach’s thought experiment overlooks is the TIME required to establish the new steady-state temperature of the outer shell, once the outer steel shell is introduced. From the moment that the outer shell is introduced until the out shell reaches it’s steady-state temperature, not all of the energy that is constantly being generated by the inner sphere is actually being radiated away from the outer surface of the outer shell. That energy, which being generated by the inner sphere but which is not being radiated away from the outer surface of the outer shell does NOT disappear. Rather, it manifests itself as kinetic energy (and thus temperature increases) in BOTH the inner sphere and the outer shell. No ‘additional’ energy is required to be created by the power-source within the inner sphere to raise the temperatures of the inner sphere and outer shell. Rather, it is only the regular energy that the power source constantly generates (but which has not been immediately radiated away from the outer surface of the outer shell), which is responsible for the temperature increases in BOTH the inner sphere and the outer shell.

To help further, let me give you an analogy, using the transfer MONEY instead of ENERGY. I know that many dislike analogies, but this one is good, so please bear with me and let the (necessarily long) story unfold….
A person has been raised to believe that they should give away 10% of their wealth (which is represented by the funds held in their checking account) to an external charity, each month.

On the first day of the first month, when this person starts their first paid job (say at a monthly rate of \$1000), they have zero dollars in their checking account, so they give away nothing to charity. At the end of “month one”, they receive their \$1000 salary paid into their checking account.

At the commencement of “month two”, they now have \$1000 in their checking account and so give away a \$100 check to charity. At the end of “month two”, they again receive their \$1000 salary paid into their checking account.

At the commencement of “month three” they now have \$1900 in their checking account and so give away a \$190 check to charity. At the end of “month three”, they again receive their \$1000 salary paid into their checking account.

At the commencement of “month four” they now have \$2710 in their checking account and so give away a \$271 check to charity. At the end of “month four”, they again receive their \$1000 salary paid into their checking account.

You will see that eventually their checking account will reach \$10,000 and they will give away \$1,000 each month to charity, and they will earn \$1,000 for the next month – the state of financial equilibrium has been reached. This is like the equilibrium state of ‘inner sphere without the outer shell in situ’ in the thought experiment.

Many years pass and this hardworking and generous person rears a child and this child is also raised with a very charitable nature. However, unlike their hardworking parent, this child (now entering adulthood) does not become an independent worker in their own right but instead, remains wholly dependent upon their generous parent for their entire income. The parent, believing that “charity begins at home”, continues to give 10% of their wealth away but now, all of their charitable giving goes to their dependent child, rather than the external charity.

On the first day of the first month (donation day), when the dependent child leaves home, the dependent child has no wealth in their own checking account and so gives no charitable donation to any external recipient and similarly gives no charitable donation to their hard-working parent either. However, the parent has \$10,000 in their checking account, so the parent now gives away 10% of their wealth (as a \$1,000 check) posted to their dependent child – and, on arrival, this check is paid into the dependant child’s own checking account, so the dependant child’s checking account jumps to \$1,000 dollars, whilst the parent’s checking account drops to \$9,000. However, by the end of the first month, the parent is paid their regular \$1,000 salary so the parent’s own wealth is again restored to \$10,000.

At the commencement of “month two” (donation day), the parent posts another \$1,000 check to the dependent child. For the first time, the dependent child similarly posts a check for 10% of it’s own wealth (the \$1,000 gratefully received last month) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent (the child, like the shell in the thought experiment, has two directions for giving). So, the child posts a check for \$100 to an external charity and posts a check for \$100 check back to their parent (leaving the child \$800). The two checks sent between the parent and the dependent child always cross in the post. On their arrival, the dependent child will now have \$1,800 whilst the Parent will now have \$9,100 but, by the end of “month two” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,100.

At the commencement of “month three” (donation day), the parent posts a \$1,010 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now \$1,800) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for \$180 to an external charity and posts a check for \$180 check back to their parent (leaving the child \$1,440). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have \$2,450 whilst the Parent will now have \$9,270 but, by the end of “month three” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,270.

At the commencement of “month four” (donation day), the parent posts a \$1,027 check to the dependent child. Similarly, the dependent child posts a check for 10% of it’s own wealth (now \$2,450) to an external charity and also posts a check for 10% of it’s own wealth back to their generous parent. So, the child posts a check for \$245 to an external charity and posts a check for \$245 check back to their parent (leaving the child \$1,960). The two checks sent between the parent and the dependent child again cross in the post. On their arrival, the dependent child will now have \$2,987 whilst the Parent will now have \$9,488 but, by the end of “month four” the parent also receives their regular \$1,000 paycheck, so the parent’s own wealth is restored, but this time to \$10,488.

You will find that, eventually, the parent’s checking account will grow to reach \$20,000 and that the parent (whilst still earning a salary of only \$1,000 per month) will be required give away \$2,000 each month to the dependent child. The dependant child’s checking account will grow to reach \$10,000 and the dependent child will give a \$1,000 check to their ‘external’ charity and will give a \$1,000 check to their parent – the new state of financial equilibrium has been reached. This is like the steady state of ‘inner sphere and the outer shell in situ’ in the thought experiment.

All the ‘additional’ money in the system is accounted for (it only ever came from the gainful employment of the working parent and yet, by the introduction of a wholly dependent child, the parent has, over time, become wealthier – twice as wealthy in fact – whilst still only earning the same \$1,000 salary each month. Neither the parent nor their dependent child has fraudulently created any fake money. The wealth held in the two checking accounts is entirely genuine – but the (wholly dependent) yet generous child’s “back-giving” has allowed the parent’s own wealth to increase.

A dollar, when given by a poorer person to a richer person, must inevitably make the richer person one dollar richer. However, the richer person will always ‘outgive’ to the poorer person – the net flow of dollars is always from the wealthier parent to the poorer child.

The same is true with the energy conveyed by the photons that are emitted from a colder object toward a warmer object – the radiant energy from these photons will be thermalized by the warmer object (like the dollar from a poor person, each Joule has to be accounted for). However, the warmer object will always ‘outgive’ the amount of energy it gives to colder object. The net flow of energy is always from the hotter surface to the colder surface.

The colder object does not prevent the hotter object from emitting all of it’s radiant exitance at the level prescribed by the S-B law. Similarly, the hotter object does not prevent the colder object from emitting all of it’s radiant exitance (on both it’s surfaces) at the level prescribed by the S-B law.

As I said earlier, ‘back radiation’ between the sphere and the shell is only significant in a vacuum. As soon as a gas is present in the gap between the sphere and the shell, the standing temperature difference between the sphere and the shell is reduced (the gas molecules act like a ‘resistive’ short-circuit). As the density of gas molecules is increased, the ‘resistive’ short circuit will tend to become a true ‘short circuit’ where the sphere temperature and the shell temperature are exactly the same.

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1. Excellent post.

One caveat: There will never be a “same temperature” under any circumstance following a radial line, precisely because of the radius difference.

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7. Steve Titcombe says:

Zoe,

Regarding the caveat: Yes, if we are splitting hairs, then you are correct: the radius of the inner sphere (and hence surface area) must always be smaller than that of the outer shell and so the Watts/M^2 radiant emittance (and hence S-B temperature) must always be greater on the surface of the inner sphere. However, in my defence, in the opening paragraph I did state that “…the thought experiment requires a vacuum within the infinitely small gap between the inner sphere and the outer shell”. That ‘infinitely small gap’ allows for the Radius of the inner shell to approximate to the Radius of the outer shell. That is important for the mathematics used in the thought experiment.

To assist you to imagine the thought experiment more faithfully, try imagining the inner shell to comprise a steel ball bearing with a 2mm diameter which is tightly surrounded by 999 steel shells each 1mm thick acting as laminations. You have now created a steel sphere of radius 1000mm. Now, imagine that you have miraculously ‘vanished’ the penultimate lamination from within the sphere and you have created an inner sphere with a radius of 998mm and a vacuum gap of 1mm and an outer steel shell which has an inner radius of 999mm and an outer radius or 1000mm. Once you’ve then miraculously inserted the radio-active power source into the centre of the of the inner sphere then you have constructed (in your head of course) a reasonable working model for the thought experiment to operate, with both the inner sphere and the outer shell each having a radius of approximately 1 metre.

If the vacuum gap was instead allowed to be filled with gaseous molecules then the inner shell would no longer represent it’s S-B temperature (being 1.189 times greater than the Kelvin temperature of the outer shell), as Radiation would no longer be the only energy transport mechanism. Instead, convection would allow more energy to move across the air gap to the outer shell and thus causing the surface temperature of the inner shell to be reduced closer to (although not exactly the same as) that of the outer shell. And the more gas molecules place in that air-gap then the more effective the convection would be. That was the point that I was trying to make. To help further: if that penultimate lamination were to be re-inserted i.e. miraculously ‘unvanished’, then the energy transport mechanism would then comprise only conduction, such that the temperature of the laminations which previously comprised the inner shell would be identical to the surface temperature of the outer surface (assuming steel has perfect conductance).

As I have said, several times, the mathematics used by Willis does require the radius of the inner sphere to be the ‘same’ as the radius of the outer shell. If the radius of the inner sphere was allowed to be 1 and the radius of the outer shell was allowed to be 2 then simple trigonometry can be used to show that only 1/3 of the ‘back-radiation’ from the inner-surface of the outer shell will strike the surface of the inner sphere such that the S-B temperature of the surface of the inner sphere will only be the fourth root of 1.3333 times greater than that of the Kelvin temperature of the outer shell i.e. the ability of the outer shell to affect the surface temperature of the inner sphere is substantially reduced.

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8. Barry Smith says:

Glad to see you’re back posting.

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9. SamH says:

Zoe, I have found a mistake in your calculations. In the third line of the section with the steel shell, you plug in 254 K for T1. However, this value (254 K) was calculated from the case without the steel shell, so it no longer applies when you add the steel shell. You have to treat T1 as a variable.

As a result of using the old value for T1, your solution with the steel shell no longer obeys energy conservation. The outer shell emits less energy to space than the nuclear reactor is producing. So energy is piling up in the system, and it will get hotter.

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1. Wrong. T1 is not and can not be a variable, for the reason I explained. Nuclear reactions are FIXED! You will not get more energy from the fission of fuel based on anything else down the line.

I’m not an expert on nuclear physics, but that is absolutely true.

As a thought experiment, take a regular 9V battery, and create a heat source by directly connecting the terminals. The max power you get is P=IV= 0.5 A × 9V = 4.5 W.

Let’s assume the battery has infinite “fuel”, voltage never drops, and you have perfect conversion to heat. You can couple the “short” to a metal shell with an inner surface area of 1m^2.

So, you will have the same experiment, but the inner metal shell gets 4.5W/m^2. If we imagine the shell to have zero thickness, it will also emit 4.5W/m^2 to space.

But let’s add some thickness. Now we got 3 W/m^2 emerging to space.

This is where the problem starts. Will adding another shell cause more than 4.5 W/m^2 to be available at the inner wall? Will the battery crank out more power? No, of course not.

The max temperature of the inner wall is limited to ~94K. It is max, it is fixed. Nothing else you put down the line will cause it to rise.

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1. ElectroNeutrino says:

While the energy density of the fuel is fixed, the rate of reaction, and therefore the rate of energy released, and therefore the equilibrium temperature, are definitely not fixed, and depend on factors such as density of the fuel, neutron flux, and even the speed of the neutrons. If the rate of reaction were fixed, then we wouldn’t be able to control the output of reactors, or make nuclear weapons, nor would be have to worry about meltdowns.

At equilibrium, the total rate of energy emitted by the outer shell must equal the total rate of energy being absorbed by the inner shell, and this rate of energy must match the power output of the reactor. while the rate of energy transport across a medium is proportional to the difference in temperature on either side and inversely proportional to the thickness.

Putting it all together, the power output is given by the reactor. The temperature of the outside part of the shell is then determined by the power output and the radius using the Stefan-Boltzmann Law. And the temperature of the inside part of the shell is given by the the power output, the outer shell temperature, and Fourier’s Law.

If the temperature of the reactor were fixed, then you have a violation of either the first or second laws of thermodynamics, since either you have energy disappearing or the reactor heating the inner part of the shell higher than its own temperature, depending on if you set the inner part no higher than the reactor, or if you have the same total emission on the outer shell as the power output of the reactor.

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1. I gave a link. The link shows diagrams with a maintained room temperature. The temperature is not adjusted to that of what’s outside. The fuel rate is adjusted to maintain the enclosed (inside building) temperature.

So whether it’s hot or cold outside, the reactor’s surface is set to room temperature. I just use that same assumption. What’s inside the steel housing is the room, and its temperature is maintained.

What would happen if the flux, rather than temperature, was maintained constant and the sink (people’s usage of generated electricity) was abruptly halted? Oopsie

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2. ElectroNeutrino says:

Which means that the power output of the reactor is adjusted to keep the temperature the same, and so doesn’t address the point that with a given power output, T1 will be different with the shell versus without.

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3. A reactor is set to keep T1 stable. It is DANGEROUS to keep the flux stable.

I think you misunderstand stability. The logic circuit is not set to make stable whatever new temperature may arise by keeping the flux the same. It is set to a specific temperature. If the room temperature in a real reactor goes to 40C for some unknown reason, that is also dangerous, and the reactor will reduce fuel further so that the wall sensor is back at 20C, or whatever is the standard.

I don’t know why that’s hard to understand. My home central heating/cooling system is set to a fixed temperature: 71F. When it gets too hot, the AC goes on. Too cold, the heat goes on. My system is not set to produce a fixed flux, and neither is a normal nuclear reactor.

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4. ElectroNeutrino says:

It’s not that it’s hard to understand, it’s that you’re not comparing apples to apples anymore when you vary the flux.

The entire point of the thought experiment is that with a given flux, T1 will change. If the flux is no longer given, then it’s a completely different thought experiment.

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5. No, the entire point of the thought experiment is that T2 will change. T1 doesn’t exist in Willis’ set up.

No existing nuclear reactor ever built creates a constant flux.

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6. ElectroNeutrino says:

Willis sets the flux at a constant amount, in his case 235 W/m^2, and lets the temperature change as determined by the Stefan-Boltzmann law, Fourier’s law, and the conservation of energy.

You vary the flux to force the temperature the same, so it’s no longer addressing the same points.

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7. Right. His nuclear reactor is a unicorn. Mine is a horse with an attached horn.

The central claim of the greenhouse gas theory is that outgoing flux is reduced. Is it not? Reducing emissivity should lead to a warming. But Willis keeps the emission the same. So who is misleading?

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8. ElectroNeutrino says:

Close; the core principle of the greenhouse effect is that the ratio of surface flux to escaping flux is reduced via absorption and emission of IR by greenhouse gasses, e.g. only a certain percentage actually passes through. It’s the Beer-Lambert law applied to the atmosphere.

At equilibrium, the first law of thermodynamics dictates that the escaping flux is equal to the incoming flux. So the outgoing flux is determined entirely by the incoming flux.

If you compare two systems with the same flux, one without the reduction and one with the reduction (either via Fourier’s law or the greenhouse effect), the latter will have a higher surface temperature than the former, just from those two above principles.

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9. lol. but you kept the albedo fixed!

You have 235 initially (current GHG levels) because albedo is set to ~0.3 from the ~340 we get from the sun.

Let’s say we introduce so much GHGs that Earth now emits 117.5.

Since the albedo is determined to be just (shortwave in – longwave out) / shortwave in

All that will happen is that albedo will increase to ~0.655.

The solar received at the surface will be ~117.5.

There’s only a warmup because you adjust some parameters and not others.

I account correctly that new fluxes will be 117.5 and 117.5, but you want to keep 235 and 235 as if it’s some sacred fixed number.

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10. ElectroNeutrino says:

Albedo is the ratio of reflected flux / incoming flux of visual (shortwave) radiation. Emissivity is the ratio of emitted thermal radiation to that of a perfect blackbody. What you described is the imbalance ratio, which is zero for equilibrium.

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11. Thought you’d go there.

“Albedo is the ratio of reflected flux / incoming flux of visual (shortwave) radiation”

Yeah, and in order to satisfy the equilbrium at TOA, reflected shortwave = incoming shortwave minus outgoing longwave.

So if you reduce outgoing LW, you increase reflected shortwave … albedo goes up.

The amount of shortwave reaching the surface is reduced.

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12. ElectroNeutrino says:

That’s not true at all. Reflected shortwave = incoming – absorbed shortwave flux. Absorbed shortwave flux is only equal to outgoing longwave flux at equilibrium. If outgoing longwave flux does not match absorbed shortwave flux, then it’s no longer in equilibrium and the temperature changes until it’s equal due to the first law of thermodynamics.

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13. Again, you keep albedo fixed as if it’s a conserved quantity. It’s not.

Now you’re suggesting that GHGs do not in fact reduce outgoing radiation. And yet the textbook math requires it.

Look at my response to Jarle.

“is only equal to outgoing longwave flux at equilibrium.”

Yeah, and a new equilbrium will be established, and there is no fixation to the old.

We’re not even talking about the real albedo – what determines surface insolation. Only 48% of insolation reaches the surface, not 70%.

On Venus, albedo is 80%! and only 1-2% of TOA insolation reaches the surface.

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14. Tom Dayton says:

Zoe wrote “So if you reduce outgoing LW, you increase reflected shortwave … albedo goes up.”

There is no physical mechanism for that to happen.

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15. Really? Filling the atmosphere with more molecules that can absorb in shortwave such as H2O and CO2 will not affect albedo? No more clouds will be created if we add water vapor galore?

Funny how Venus has 80% albedo, and only 1-2% insolation reaches the surface …

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16. Tom Dayton says:

CO2 has an inconsequential absorption of shortwave. Clouds are liquid water, not water vapor. Water vapor increases as a feedback to increased temperature. Increased water vapor’s major and immediate effect is absorption of longwave. Increased water vapor does allow for increased clouds, but different kinds of clouds have differing balances of effects on longwave retention and shortwave reflection. Any effects on shortwave albedo are distinctly secondary in time, causality, and influence.

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17. And yet … while only 48% of the sun reaches Earth’s surface, only 1-2% reaches Venus’ surface.

CO2’s absorption bands and transmissivity are also set by temperature and pressure. Venus has an incredible new set of absorption bands that are not seen on Earth, including in the shortwave portion.

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18. ElectroNeutrino says:

It is completely possible to reduce the longwave emission without touching albedo, e.g. with CO2 alone. Water vapor concentration is entirely dependent on the Clausius–Clapeyron relation.

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19. Right now at TOA:

340 SW IN = 105 SW OUT + 235 LW OUT

Now we reduce LW out to 230

340 > 105 + 230

Imbalance. Now what happens?

Don’t contradict yourself a 2nd time.

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20. ElectroNeutrino says:

“If outgoing longwave flux does not match absorbed shortwave flux, then it’s no longer in equilibrium and the temperature changes until it’s equal due to the first law of thermodynamics.”

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21. Be more clear. We’re not talking about temperature changes at the bottom. We are talking about TOA fluxes. Are you saying that 230 LW returns to 235?

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22. Tom Dayton says:

Zoe asked “Now we reduce LW out to 230. 340 > 105 + 230. Imbalance. Now what happens?”

Energy accumulates in the atmosphere–lower in the atmosphere first, because there is more LW absorption happening by the air above there. LW heading toward space is intercepted by greenhouse gases between it and space, then transferred by collision to surrounding molecules of all types, and some of it reradiated in all directions. The now-warmer greenhouse gases increase their radiation because they have more energy.

That’s more outgoing radiation trying to squeeze through the obstacle course of greenhouse gases between those radiating molecules and space. Failing to squeeze through, that extra radiation is absorbed by those intermediate molecules, thereby increasing the energy in those intermediate molecules.

The proportion of LW radiated toward space, that actually makes it to space without being intercepted, switches location from being radiated by greenhouse gases lower in the atmosphere, toward those higher in the atmosphere. But those gases higher in the atmosphere radiate less than the gases lower in the atmosphere, because the former are colder. That lesser radiation makes those higher gases retain more absorbed energy, thereby warming.

Eventually the greenhouse gases in the lower layers radiate enough, so even with their difficulty of making it all the way to space, enough LW does make it to space to for their outflow to balance the inflow from the Sun.

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23. Skip the drama. Why can’t you just give a simple answer?

Let’s try again:

340 > 105 + 230

Imbalance. Now what happens?

340 = 105 + 235

YES or NO

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24. ElectroNeutrino says:

The first law of thermodynamics dictates that the temperature will increase, thereby increasing the emitted surface flux and so the TOA flux, until the TOA flux matches the incoming absorbed flux.

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25. OMG. Does LW return back to 235 from 230? YES or NO?

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26. ElectroNeutrino says:

Yes. The surface temperature increases as required by the first law of thermodynamics, du = dq – dw. Since the surface emitted flux is proportional to T^4, it will also increase, and since the TOA flux is proportional to the surface flux, it will also increase, until flux-in = flux-out.

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Physics works in real time. We don’t need to wait for a change in 5 W/m^2. Same principle should work for a change in 0.0000000000000000001 W/m^2.

So, how can you say it’s possible to lower OLR when it will instantaneously return to its previous state?

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28. ElectroNeutrino says:

Nobody said anything about instantaneous. That first law of thermodynamics is a differential equation, du = dq – dw where du is the differential of internal energy, dq the differential of heat, and dw the differential of work.

Since no work is being done on the atmosphere, dw = 0, so du = dq.

For most materials, like the air, u = CT, so du = CdT, where C is the specific heat capacity. And since dq is the total heat flow, dq = (P_absorbed – P_emitted) dt = (P_absorbed – (1/λ) εσT^4) dt, where A εσT^4 is the Stefan-Boltzmann law for a greybody, and λ is the ratio of surface emission to TOA flux (e.g. the strength of the greenhouse effect).

This means that the rate of change of the temperature, dT/dt = (P_absorbed – (1/λ) εσT^4) / C.

Equilibrium occurs when dT = 0, which happens when P_absorbed = (1/λ) εσT^4, rearranging gives T = (λ*P_absorbed/εσ)^(1/4) as the equilibrium temperature.

With a gradual change in λ, there is a gradual change in T.

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29. Why do you have to be so obtuse? Can you stick to just one thing? We can discuss temperature later.

You said that it’s possible to reduce OLR, and at the same time claim it will be restored. So any minute change is quickly offset.

Bringing in heat capacity doesn’t change this either because it’s active in both OLR reduction and subsequent return. So you can just drop it from your argument.

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30. ElectroNeutrino says:

I’m giving you the physics and explaining the mechanism behind the outgoing flux eventually matching the incoming flux, and why it’s not instantaneous.

It takes time for the flux to match, as described in the math above, so a faster change in λ leads to a greater imbalance between incoming and outgoing flux. If you change λ fast enough, you increase the imbalance faster than it can equalize.

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31. “If you change λ fast enough, you increase the imbalance faster than it can equalize.”

Very much false. It takes same time to unbalance as rebalance.

Why? You explained yourself: The IR-active gases are surrounded by non-IR active gases. While IR active gases can drop temperature (and emission) faster, they are still surrounded by gases that can’t. Those non-IR gases will cool slower. They can only effectively cool by bumping into IR gases. And guess what? They will transfer their energy to the IR gas to emit, thereby slowing down the IR gas cooling.

It’s quite amusing that you only consider heat capacity one way but not the other way.

Heat capacity works both ways exactly the same. Which is why I urged you to drop this.

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32. ElectroNeutrino says:

That doesn’t follow, since λ is determined entirely by Schwarzschild’s equation.

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33. Uhm, what does Schwarzchild have to do with this?

Emission equations can only deal with IR active substances.

You want to warm nitrogen and oxygen by conduction on the rebalancing part, but can’t comprehend nitrogen and oxygen warming H2O/CO2 by conduction when they cool too fast on the imbalancing side.

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34. ElectroNeutrino says:

Schwarzschild’s equation gives you the transmittance of a reactive medium. dI(v) = nσ(v) (B(T,v) – I(v)) ds, where I(v) is the intensity of incident radiation at frequency v, n is the density of the reactive medium, σ(v) is the cross-section at frequency v, B(T,v) is the Planck emission at temperature T and frequency v, and s is the path.

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35. We are using bulk parameters. Why did you change topics to transmissions at wavelengths?

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36. ElectroNeutrino says:

Schwarzschild’s equation is given in terms of spectral radiance. You can simply integrate over the relevant range if need be to get the transmittance in a band, including the entire spectrum.

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37. ElectroNeutrino says:

No.

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10. SamH says:

It seems you are making an energy balance argument, but instead of calculating the difference between energy in and out, you are just stating that T1 can’t change because of energy balance.

But let’s do the math. In your first calculation, you correctly set that the energy produced by the nuclear reactor must equal the energy leaving from r1. by conduction, q=4 pi k r1 r2 (T1-T2)/(r2-r1) = 235 W.

In the second calculation, you no longer set this condition. Instead, you arrive at T2=253.884 K, and you set T1=254.041 K . Now if we calculate conduction, we get q=4 pi k r1 r2 (T1-T2)/(r2-r1) = 117.5 W. This is exactly half as much as before. But the nuclear reactor is still producing 235 W.

So the reactor is still producing 235 W, but only 117.5 W is escaping from radius r1 by conduction. Energy balance is not satisfied.

Liked by 1 person

11. SamH says:

Zoe, just to make sure we’re on the same page: which of these points do you agree with?

1) the reactor produces a constant output power, regardless of the environment, as you have just said. This is 235 W

2) in your second calculation, only 117.5 W leave the reactor through conduction

3) the reactor is producing more energy than it is getting rid of, so it must be getting hotter

Liked by 1 person

1. As I said, the temperature goes up by 0.158K.

The reactor is not producing more energy than it is getting rid of, it is producing the same energy that it is not getting rid of a little bit more.

You have a different solution? What is it?

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2. Oh I see what you’re saying. It’s simple: conduction is a loss. Your nuclear reactor is split between powering the local matter and emitting to space.

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12. SamH says:

You didn’t treat conduction as a loss in the first calculation (no shell), so why do so in the second?

In my solution to the second calculation, I have almost the same equations as you, but I do exactly what you did in the first calculation and set the energy produced by the reactor equal to the conduction away. Then your first line looks like

235 W=4 pi k r1 r2 (T1-T2)/(r2-r1) = 4 pi r2^2 sigma (T2^4-T3^4) = 4 pi r2^2 sigma T3^4

T3 = (235 W/m^2/sigma)^.25
T2 = 2^.25 T3
T1 = T2 + (235 W) (r2-r1) / (4 pi k r1 r2)

From here just plug numbers into the first equation, to get T3 = 253.73 K

Plug this into the second equation to get T2 = 301.73 K

Plug this into the first equation to get T1 = 302.05 K

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1. If you’re that clever why did you fall for my trick?

I was describing a real world nuclear reactor, but you believed my false description of a fake reactor.

https://www.nuclear-power.com/nuclear-power-plant/nuclear-reactor/

“Almost all of the current reactors built to date use thermal neutrons to sustain the chain reaction. These reactors contain neutron moderator that slows neutrons from fission until their kinetic energy is more or less in thermal equilibrium with the atoms (E < 1 eV) in the system."

In real world reactors, the flow is actually cut off to obtain equibrium temperature of environment.

The first equations tell what the operators would set as that equilbrium temperature to.

O

The other type of reactor is a fast reactor. In this scenario the flow increases.

So you see, the trick is that Willis used a power source that doesn't actually produce a constant flux 🙂

Then I distracted you with the battery example. But that would actually work as you said. But you didn't catch on.

I like to that you discovered that I'm a tricky gal. There are a few other little physics/description tricks on this site.

I never however manipulate data. That is 100% as is. If I made a calculation mistake, then it's a mistake. I provide the code to catch it. I don't play around in this area.

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13. SamH says:

Nice trick. My calculation used a constant power source, because that was the original problem statement from Willis, and more importantly, that is a more analogous situation to the earth. The amount of sunlight incident on the earth does not change based on the earth’s temperature.

But if you are assuming the power source adjusts its power to maintain a constant temperature, regardless of the environment, of course the result is that you will get a constant temperature. In that case your calculation may be more appropriate.

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1. Yes, correct. Willis stated a constant power source, but used a physically unconstant power source.

T1 is constant, but T2 changes, with a normal nuclear reactor.

I’m not assuming discarding the environment. The environment IS the inner wall of steel housing. The designers made the device work as initial equations show.

I’m glad you took the bait. I like discussions. Thank you very much!

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2. SamH says:

Always good to have a thoughtful discussion, thanks.

I think the next natural question is: tricks aside, let’s say we actually have a constant power source, for example a heating resistor with controlled P=I*V. Would you then agree with my solution?

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1. Yes, it’s correct. Although I don’t fully understand the electrical ramifications of having two metals separated by an infinitely tiny gap. Probably a complication there. But otherwise, yes.

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14. Joshua says:

I might be confused, but isn’t Q measured in Watts, and the flux of 235 is W/m^2? So for calculating T1, shouldn’t you be setting the left hand side of the equation equal to 235 x the surface area of the outer sphere? And would this not significantly increase T1?

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15. Jarle says:

Actually, if you look closely at this supposedly scientifically accurate animation, the temperature at the point of emission stays the same, which implies unchanging emissivity??

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1. It doesn’t make sense to me. I’d have to see the math.

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2. The basic one layer model warm-up formula is:

S(1-a)/(1-o/2)

Where S is ~340, a = 0.3, and o is opacity (1 minus emissivity) = 0.78

So the sun’s 340 W/m^2 becomes ~390 at the surface.

If opacity is increased from 0.78 to 0.81 (emissivity decreased by 0.03)

The result becomes 400 W/m^2 at the surface.

Well, that’s the basic textbook theory anyway.

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16. Alex says:

All you’ve done is succeeded in showing that you did not understand the problem set up by Willis. The internal source is intended to supply a constant energy flux. Willis’ solution to the problem is correct.

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1. But his power source is a unicorn – doesn’t exist. I use a real nuclear reactor.

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1. Alex says:

That’s not *his* power source. That’s YOUR straw man. Did you even read what he wrote?

“For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square metre of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.”

He tells you to heat the interior in any manner that you like so long as it supplies a constant 235 watts per square metre of surface area.

You didn’t do that.

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1. Exactly. His power supply is a unicorn. I use a real power supply.

He needs the output of a brown dwarf. But he doesn’t use a brown dwarf:

“For our thought experiment, imagine a planet the size of the Earth”

He needs something at least 10x the size of Jupiter.

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17. Alex says:

It’s a thought experiment. You did not satisfy the constraints of the thought experiment which was to use a power supply that provides a constant 235 W/m^2. You did not solve the stated problem correctly. Now you are stomping your feet like a child because you got called on it.

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1. Got called on what? I specifically said that the temperature is fixed. I put the world “real” in the title. If you don’t like what I did, that’s your problem, not mine.

I don’t believe in unicorns or brown dwarfs the size of our planet.

The stated problem is unreal and physically impossible.

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1. Alex says:

“I specifically said that the temperature is fixed.” That is NOT the problem that Willis described. Why is that so difficult for you to understand?

“I put the world “real” in the title.” So you think that there is a “real” nuclear reactor that can supply 235 W/m^2 to an Earth-sized sphere?

“I don’t believe in unicorns or brown dwarfs the size of our planet.” Sure you do. You believe that there is a nuclear reactor that can supply 235 W/m^2 to an Earth-sized sphere.

“The stated problem is unreal and physically impossible.” Why are you having so much trouble understanding what a thought experiment is?

Face it, your post is full of ignorance and nonsense.

Oh and by the way, conduction is not a “loss”. At steady state the net flux of energy out through any spherical surface containing the source must be 235 W/m^2.

You can admit that you don’t really understand physics.

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1. If I don’t understand physics, how did I get the right answer for a real world nuclear reactor?

“You believe that there is a nuclear reactor that can supply 235 W/m^2 to an Earth-sized sphere.”

Don’t slander me. The sphere I use has a surface area of 1 square meter.

“Why are you having so much trouble understanding what a thought experiment is?”

I don’t. My post is a thought experiment. I didn’t actually build this device.

Willis’ setup is science fiction, mine is REAL.

Did you know that real satellites don’t use “insights” from Einstein’s thought experiments? Why? Because they don’t correspond with reality. They end up fixing his insights with fudge factors, or just ignoring it all together and using parameters from actual observations.

“conduction is not a “loss”. At steady state”

I don’t use steady flux. Conduction is a loss.

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2. Tom Dayton says:

Zoe, a well known example of a thought experiment is Albert Einstein imagining himself chasing a beam of light. It is intentionally unrealistic, for the very purpose of being valid and useful. Here is an explanation: https://en.wikipedia.org/wiki/Thought_experiment

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1. But Einstein’s insights were not useful. Fudge factors need to be added to line up with observations.

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2. Tom Dayton says:

“But Einstein’s insights were not useful,” Zoe Phin claimed. Here endeth my attempt to find any value in Zoe Phin’s opinions.

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3. Because he’s a sacred cow to you? Well, Robert J Oppenheimer thought he was a useless imbecile, and he built a nuclear bomb. What did Einstein build? Thought experiments? Tell me who uses them in REAL things. I don’t mean those that are now hunting for dark matter and the like using his presumptions.

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4. ElectroNeutrino says:

“Einstein was a physicist, a natural philosopher, the greatest of our time.”

– Robert Oppenheimer, “On Albert Einstein”, December 13, 1965

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5. Oppenheimer only started to like and be friends with Einstein in the last decade of his life, mainly because they were both outspoken about the proliferation of nuclear weapons. That’s a good thing.

But by praising Einstein he gets to partially blame him too.

“Yeah uhum sure … if it wasn’t for Einstein … there would be no bomb. It was his special insights that made it possible”.

That’s a paraphrasing of what he said.

Even though Einstein didn’t think nuclear weapons were possible, and he was of ZERO use to the Manahattan project at the time.

Learn to read between the lines!

Although, Einstein wasn’t alone in doubting nuclear energy. Tesla also failed to understand it.

For Oppenheimer’s praise to be worth something, he would have to establish what great technology or LASTING insight came from Einstein that couldn’t be achieved otherwise. Just buttering up his legacy with Hallmark card rhetoric won’t do.

We know there’s very little insights that will remain standing, and a complete collapse of his paradigms is eminent. Though the mythology may still persist for now.

But to his credit, he did incite a lot of good debates, and being wrong is still scientifically useful.

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6. ElectroNeutrino says:

“His part was that of creating an intellectual revolution, and discovering more than any scientist of our time how profound were the errors made by men before then. He did write a letter to Roosevelt about atomic energy. I think this was in part his agony at the evil of the Nazis, in part not wanting to harm any one in any way; but I ought to report that that letter had very little effect, and that Einstein himself is really not answerable for all that came later. I believe he so understood it himself.”

-Ibid

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7. More along the same theme. You see how his rhetoric is shaped by his friendship and alliance to the same cause?

This was of course great, but nothing to do with Einstein’s mostly plagiarized hypotheses that made him famous and gave him a Nobel Prize.

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8. ElectroNeutrino says:

Oppenheimer explicitly said in that speech that he didn’t think Einstein deserved any of the blame, nor did he deserve any credit for it. I was directly addressing your claims that Oppenheimer blamed Einstein, even partially.

Also, Einstein didn’t get the Nobel for either special or general relativity, and it’s entirely beside the point that a thought experiment need not be physically realistic to give insight.

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9. OK, in THAT speech, sure.

The Nobel citation reads that Einstein is honoured for “services to theoretical physics, and especially for his discovery of the law of the photoelectric effect”

But he didn’t discover the photoelectric effect (Hertz). Nor was he first to suggest light is discrete packets of energy (Planck). Nor did he coin the term “photon” (Compton).

Einstein never did any physical experiments to discover anything.

Many people feel this refererence to theoretical physics is relativity by the backdoor.

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10. ElectroNeutrino says:

The photoelectric effect was seen as a paradox until Einstein was able to explain it using Planck’s theory of quantization (which, incidentally, Planck never performed any experiments either, neither did Maxwell, Dirac, Schrödinger, Boltzmann, etc.; that’s what theoretical physics is.). Nobody else thought to do so.

This explanation resolved more than just the paradox, and was able to make testable predictions on the nature of the photoelectric effect which were later confirmed. Not only that, but his explanation was one of the core principles to the formation of quantum mechanics.

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11. Whatever. Einstein is nothing without Planck and Millikan.

I use Planck’s formula to convert a spectrum into a power flux. Einstein didn’t need to exist for me to do that.

It’s a shame they gave a prize to Einstein before Millikan.

The debate in 1921 was whether to have ZERO prize or give it to Einstein. It’s obvious it was more of a politcal move, as usually there are several contenders every year.

There was no debate between which candidate to pick, and in a haste they gave to Einstein.

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12. ElectroNeutrino says:

Even so, he didn’t get the Nobel prize for either theory of relativity.

And Planck is nothing without Stefan or Boltzmann. And both of those would be nothing without Maxwell, Clausius, And they would be nothing without Ampere, Faraday, Gauss, etc. And so on and so on. Even Newton and Huygens would be nothing without Copernicus, Brahe, Kepler, Galileo, …

Physics, and indeed science in general, is built on the improvement and extension of the work of those that have come before.

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13. But Planck offered something new. He proposed E=hf.

Once you offer something like that, it’s obvious. Quanta is already in the formula.

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14. ElectroNeutrino says:

And Einstein proposed K = hf – W for the photoelectric effect, which was later verified. Like I said, he was the one that resolved a major paradox of the time.

He was also the first to propose that Planck’s E-hf meant that light always came in discrete packets. Before then, everyone thought it was just a mathematic trick to make the equation give the right values. Using that proposition, he explained why photoelectric effect didn’t obey Maxwell’s equations. His explanation was rejected for years because it ran contrary to the prevailing idea of the theory of light radiation, until Millikan verified it experimentally, which he did so only because Einstein proposed it.

He also proposed Ruv – 1/2 R guv = Tuv, which is probably one of the most revolutionary equations in physics.

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15. There are wave-only theorists that have no problem explaining light with just waves.

Isn’t K = hf – W just the first law of thermodynamics with different letters and Q=hf ?

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16. ElectroNeutrino says:

The wave-only attempt at explaining the photoelectric effect is exactly what lead to the paradox that Einstein resolved.

K isn’t the internal energy of a system, it’s the kinetic energy of a photoelectron ejected by a quantum of light out of a potential of W; though it is inspired by the first law of thermo. That’s why W is called the work function, even though no work is actually being done.

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17. There are wave-only theorists today! They don’t see any paradox. The “paradox” is a created narrative.

1st LoT doesn’t have “internal energy of a system”. It has a CHANGE of internal energy of a system. An ejected photon would cause a change.

It really is just a restatement of 1st LoT. Not impressed at all.

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18. ElectroNeutrino says:

Can you cite an example of one of them explaining the photoelectric effect with wave-only physics?

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19. No, because I don’t have a link to everything I read. But I can tell you that if you assume it exists and are open minded … you will easily find it.

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20. ElectroNeutrino says:

That’s not how science works, and leaves one open to believing in things that don’t actually exist except in the minds of those that want it to.

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18. Jarle says:

Einstein deserves credit. However, the glorification of him is probably due to a very one-sided promotion of him. It is always negative when such a “climate” arises. It is just as forbidden to criticize Einstein as it is to criticize negative consequences of immigration or negative effects of transitioning to renewables. Do it, any you are instantly turned into a fool that nobody wants to have a discussion with anymore.

I found this book a delight. A balancing weight to the mad glorification af Einstein.

Liked by 2 people

1. Of course he deserves credit. Being wrong about a lot of things tells others what not do.

I’m not right about a lot of things either, but I don’t publish papers and persist that they’re right – causing others to chase phantoms. Nor do I hire/persuade people to slander the experiments of others.

At least he confessed in private that he worries he may have achieved nothing that will last.

“Do it, any you are instantly turned into a fool that nobody wants to have a discussion with anymore.”

Ha! I get that all the time in my professional work discussing things with non-experts and experts that will be shown to be wrong.

There’s a parameter in HITRAN’s database named after Einstein. So maybe he did something right. Or he plagiarized that too?

He was certainly great at self-promotion and calling his critics “racist” when he couldn’t address their reasonable arguments.

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19. Alex says:

“If I don’t understand physics, how did I get the right answer for a real world nuclear reactor?”

You didn’t. A “real-world” nuclear reactor cannot maintain a perfectly constant temperature.

“The sphere I use has a surface area of 1 square meter.” – And of course that is not Willis’s problem statement. Nor is your straw man that you can use a real-world nuclear reactor to satisfy his constant energy source parameter.

“My post is a thought experiment.”

Your post claims that Willis’s solution to his problem is wrong. It is not, hence your post is wrong. Face it, you either did not understand the parameters set up in the problem, or you purposefully lied about them.

“Willis’ setup is science fiction, mine is REAL.”

Are you sure? What “real-world” nuclear reactor will fit inside a sphere with 1m^2 of surface area? What object has an emissivity of 1? Keep digging that hole.

“Conduction is a loss.” Nope. Conduction does not annihilate energy. It is not a loss.

“I don’t use steady flux.” Are you sure? If you are not calculating steady state, then where is your time derivative term?

Face it. You don’t understand basic physics.

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1. ‘A “real-world” nuclear reactor cannot maintain a perfectly constant temperature.’

Sure, it can’t keep it exactly perfect, but that’s the aim!

https://www.nuclear-power.com/nuclear-power-plant/nuclear-reactor/

‘What “real-world” nuclear reactor will fit inside a sphere with 1m^2 of surface area.’

The kind where all energy is made to generate heat, not electricity.

Ever shorted a battery? Gets hot, doesn’t it? Extrapolate to nuclear …

‘Nope. Conduction does not annihilate energy. It is not a loss.’

That’s not what a loss means, bozo. It means the external flux is reduced. That’s why we insulate things. The energy is not annihilated, it goes into the insulating material.

‘If you are not calculating steady state, then where is your time derivative term?’

Steady state and steady flux are not the same thing. Steady state is when there is no further temperature changes. Steady flux is when the flux doesn’t change. The two can overlap depending on your problem, but they are not the same.

“What object has an emissivity of 1?”

That is just such a petty argument. Don’t give yourself away so easily.

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1. Alex says:

“The kind where all energy is made to generate heat, not electricity.
Ever shorted a battery? Gets hot, doesn’t it? Extrapolate to nuclear ”

Are you not aware that in order to maintain a constant temperature a control system is needed in order to control the reaction and thus the heat output?

What “real world” control system are you using that fits inside a sphere with 1 m^2 of surface area?

“That’s not what a loss means, bozo. It means the external flux is reduced.”

If the external flux is reduced, then where has the energy gone?

“The energy is not annihilated, it goes into the insulating material.”

And at steady state, the same amount of energy that goes into the material, goes out. You don’t understand what insulation does. It changes the temperature gradient required to pass the same amount of heat.

“Steady state is when there is no further temperature changes.”

And that is what you calculated. You are so clueless that you don’t even understand the calculation that you performed. You have no time-dependence in your calculations and thus you are attempting to calculate steady state conditions.

“That is just such a petty argument.”

The pettiness has been set up by you from the beginning. You aren’t calculating anything “real-world”. You didn’t understand the specifications of Willis’s problem and now you are squirming to cover your ass.

You don’t understand basic physics.

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1. Do you even read the material given to you?

It’s not just for a steady temperature that you need a control mechanism for. You need to cut the fuel to prevent a critical state, and so the story is the same for a constant flux – which is never done.

You do realize that if citizens do not use the electricity made by nuclear plant … it doesn’t continue to “pump” the same energy. You understand what would happen if a nuclear plant produced a constant flux while demand dropped?

Do you even have a brain?

As for the control mechanism, it’s not a problem. There’s a company that designed and bringing online reactors that produce 60MW and take up [much[ less than 1000 cubic meters for the entire operating chain. Can you math? Well, maybe it doesn’t scale perfectly, but heyo, we couldn’t build tiny drones decades ago. So there is little doubt about the feasibility.

“You don’t understand what insulation does. ”
You don’t know what words mean, since you interpret them to your own liking. If you don’t like the word “loss”, how about “sink”? Have you heard of a heat sink?

Do you put a heat sink on your computer processor to warm or cool it? Think hard about this. Times up … you definitely want a a good heat sink to cool your processor.

The term “sink” was precisely created as an analogy to what happens in a sink: water falls down the drain, out of sight, and reduces the water content in your sink. The water is “lost” from your POV.

‘You are so clueless that you don’t even understand the calculation that you performed.
You have no time-dependence in your calculations and thus you are attempting to calculate steady state conditions.’

You are so incredibly obtuse. Yes! I’m doing steady state without a constant flux. You are more than welcome to run this setup through time, and you will get the same result.

The temperature in a nuclear reactor is CONTROLLED. Fuel flow is reduced.

As soon as you raised dT/dt your dq/dt is reduced, you uneducable parrot.

“You didn’t understand the specifications of Willis’s problem.”

I very much understood the specifications of Willis problem. That’s why you accused me first of a strawman, remember? Be consistent in your accusations, jerk face. I already explained to someone else that I can be tricky, and modeled a REAL WORLD scenario.

Call it a bait and switch if you will, but it got good engagment. Much better than usual.

“You don’t understand basic physics”

I’m glad you understand the physics of a real world nuclear reactor. You should apply for head of a nuclear plant with your knowledge that flux must be kept constant without regard to plant temperature and customer demand.

Now take the time to figure out why nuclear techs would not want to build a “constant flux” reactor in space. Seriously, give it some good thought.

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20. Alex says:

“It’s not just for a steady temperature that you need a control mechanism for. You need to cut the fuel to prevent a critical state, and so the story is the same for a constant flux – which is never done.”

I’m not the one that has a problem with the need for a control mechanism. YOU do. There are NO real-world nuclear reactors that can fit inside a sphere with 1m^2 surface area. NONE. YOUR solution is make-believe. It’s not real-world in any way, shape, or form.

“You do realize that if citizens do not use the electricity made by nuclear plant … it doesn’t continue to “pump” the same energy. You understand what would happen if a nuclear plant produced a constant flux while demand dropped?”

Entirely irrelevant to Willis’s problem. The problem statement REQUIRED a constant energy source. YOU did not solve the stated problem. You made up some other make-believe problem.

“Do you even have a brain?” Clearly my brain works far better than yours. If you had a reasonable IQ you would have recognized how ludicrous your arguments are and have given up by now.

“There’s a company that designed and bringing online reactors that produce 60MW and take up [much[ less than 1000 cubic meters for the entire operating chain.”

Name it and show that ALL dimensions will fit inside a 1m^2 surface area sphere, including ALL fuel rods, and control system. Then you can go ahead and explain how this REAL-WORLD system can operate while entirely enclosed in a perfect blackbody spherical shell.

“Have you heard of a heat sink?” Indeed I have. I also know that it has nothing to do with your solution. Your solution has no time dependence and thus it is a solution for steady state. This should be obvious to anyone that understands basic heat transfer.

“Yes! I’m doing steady state without a constant flux.” WRONG. The flux is IDENTICAL through any surface that encloses the heat source. You REDUCED the heat output of the source fore the problem with the surrounding shell thus VIOLATING the conditions put forth in Willis’s problem.

“The temperature in a nuclear reactor is CONTROLLED. Fuel flow is reduced.” In YOUR make-believe problem that may be the case, but in the problem that Willis proposed the HEAT OUPUT of the source is controlled, thus fuel flow is NOT reduced. Are you so obtuse that you don’t think a control system can be fabricated to control heat output? If you were as smart as you think you are then you would have come up with a real-world system that DOES satisfy Willis’s problem statement instead of one that DOES NOT.

“As soon as you raised dT/dt your dq/dt is reduced,” Not is you wanted to solve the ACTUAL problem that Willis posed. Instead you made up some different make-believe problem. Why is that?

“I very much understood the specifications of Willis problem, you uneducable parrot.” Then why didn’t you set up a real-world system that satisfied his problem statement you incompetent ignoramus?

“That’s why you accused me first of a strawman, remember?” I’m STILL accusing you of a straw man. That’s all you have because you weren’t smart enough to come up with a real-world system that is consistent with Willis’s problem statement.

“I already explained to someone else that I can be tricky, and modeled a REAL WORLD scenario.” Tricky? You aren’t even intelligent enough to understand that you haven’t solved a real-world problem, and you aren’t smart enough to figure out a real-world system that is consistent with the actual problem statement.

“Call it a bait and switch if you will” Actually, I call it sheer stupidity and now you are trying to cover your ass.

“I’m glad you understand the physics of a real world nuclear reactor.” You keep using that phrase “real-world” but you don’t even have a clue about how real-world reactors work.

“Now take the time to figure out why nuclear techs would not want to build a “constant flux” reactor in space.”

It makes no difference if they want to build it or not. Willis’s problem requires a constant power source. He never said it had to be a nuclear reactor. You really are logically challenged. Keep digging that hole though. It’s getting deeper.

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1. The parrot passed the age where it can learn new phrases.

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1. Alex says:

Good one Zoe. I also like to eat feces.

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21. Alex says:

“There’s a company that designed and bringing online reactors that produce 60MW and take up [much[ less than 1000 cubic meters for the entire operating chain. Can you math?”

I can’t believe I let that pass. I assumed that you were intelligent enough to make at least one valid point. So you’ve now admitted that there is no real-world nuclear reactor that will fit inside a sphere with surface area of 1m^2.

So, you can dispence with your “real-world” nonsense now.

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1. In my experiment:
r1 = 0.282 × 0.75 = 0.2115

V = 4/3*pi*r^3 = 0.0396 m^3

Nuclear reactor:
60MW / 1000 m^3 = 60KW/m^3

Scaling:
60 KW / m^3 × 0.0396 m^3 = 2376 W

2376 W > 235 W QED.

You can’t even do basic math.

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1. Alex says:

You’re right, Zoe. Now I see that I’m a moron.

We’ve been able to build toy helicopters, toy race cars, and basic logic should extend to us being able to build nuclear reactors inside smaller objects.

You scaled linearly and still had a 10x buffer left over. That is impressive.

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2. Alex says:

Aww, poor Zoe. I’ve been more abusive than anyone in the history of your blog. I’m sorry, but I can’t control myself. You may as well go ahead have your first ban.

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22. Alex says:

I see that you couldn’t bear being shown to be wrong Zoe. That’s OK. It’s a common trait of dishonest deniers like you that suffer from the Dunning-Kruger effect. Enjoy your blissful ignorance.

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1. Regardless of things being explained to you repeatedly, you ignore it, and continue to be an obnoxious abusive ass.

“Har har har … nuclear doesn’t scale linearly … I don’t actually know how it scales, and I’ll also ignore the 10x buffer you left har har har you ignorant Zoe.”

“You should’ve done exactly what Willis did … imagine a planet the size of Earth that can emit like a brown dwarf >10x the size of Jupiter har har har because that is more realistic than imagining innovation in nuclear physics”

“I hate you for having different criteria than Willis. You were supposed to just rewrite his post in your own words, because that is interesting.”

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1. Alex says:

“Regardless of things being explained to you repeatedly, you ignore it”

You’re describing yourself Zoe. We’ve now established that there is no “real-world” nuclear reactor that can fit inside a sphere of the size your analyzed.

I don’t hate you Zoe. I pity you. You think that you have uncovered something profound when all you have done is to show that you don’t understand basic science.

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1. “We’ve now established”

Have we? The question is if it could be done, not whether it has.

YOU JUST KEEP REPEATING that it’s not possible. Evidence? Nothing but your say so.

“you don’t understand basic science.”

Classic loser bully tactic.

You still didn’t get a different answer than me under my conditions.

You really still believe someone would be crazy enough to make a fixed flux nuclear reactor.

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23. Max Polo says:

I have to admit this was entertaining: I clearly came in late to check Zoe’s math. But why do scientists (either amateur or not) insist on thought experiments? Isn’t physics based on real experiments? I came across this recent article by Hermann Harde (https://scc.klimarealistene.com/produkt/verification-of-the-greenhouse-effect-in-the-laboratory/) who seem to have done a proper experiment that shows that greenhouse effect “works as advertised” (although he admits there’s nothing to be worried about…). I am quite puzzled because in my small way I did some tests (at constant flux !…so respecting Willis’ requirement) that showed “no back-radiation heating” (this was discussed a bit in the comments at https://phzoe.com/2020/03/04/dumbest-math-theory-ever/). I was happy with my result…but now it seems there are new data of much better quality than mine, that tell the opposite…

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1. Now that I had some time to look through it, I noticed this is not the full paper. I considered purchasing it, but I noticed something strange that stopped me: Do they have TWO power inputs?

This setup is too complicated to make sense of. I think I can, but the whole thing smells funny to even bother. Sorry

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24. Max Polo says:

Dumb question : why is not the solution of the nuclear core @ r2 + the shell @ r3 as follows: T2 = 253.73 K (dictated by the imposed flux 235 W/m2), T3 = 213.36 K (= T2/2^0.25) ? Total emitted power to space = 235 W (core) – 117.5 W (shell towards the inside) + 117.5 (shell towards the outside) = 235 W = as imposed.

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25. Max Polo says:

OK, then let’s say that I’m just about right (0.13°C difference for T3). For us engineers such a difference is invisible 🙂

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